# ncert solutions for class 8 maths chapter 2 linear equations in one variable in English | class 8 chapter 2 maths exercise 2.1, 2.2, 2.3

Students can practice the questions based on the same through class 8th maths ncert solutions for class 8 maths chapter 2 exercise 2.1, ncert solutions for class 8 maths chapter 2 exercise 2.2, ncert solutions for class 8 maths chapter 2 exercise 2.3 and many more. These examples are in the form of word problems.

The next topic is Solving equations having the variable on both sides. In this section, students will learn how to solve such equations which have expressions with the variable on both sides. class 8th maths

Here we solve ncert solutions for ncert maths class 8 chapter 2 solution From class 8 maths chapter 2 pdf concepts all solutions with easy method with expert solutions. chapter 2 maths class 8 help students in their study, home work and preparing for exam. Soon we provide CBSE chapter 2 maths class 8 solutions English Medium. NCERT solutions for class 8 maths chapter 2 pdf Linear Equations in One Variable PDF here. You can download ncert english book class 8 from official NCERT website or** Click HERE**.

# ncert solutions for class 8 maths chapter 2 linear equations in one variable in English

class 8th maths

## ncert solutions for class 8 maths chapter 2 Linear Equations in One Variable in English

### NCERT solutions for class 8 maths chapter 2 pdf┬а– Questions Answer Click Here

## ncert solutions for class 8 maths chapter 2 EXERCISE 2.1 class 8th maths

**Solve the following equations.**

1. ? тАУ 2 = 7 | 2. ? + 3 = 10 | 3. 6 = ? + 2 |

4. \(\displaystyle \frac{3}{7}+x=\frac{{17}}{7}\) | 5. 6? = 12 | 6. \(\displaystyle \frac{t}{5}=10\) |

7. \(\displaystyle \frac{{2x}}{3}=18\) | 8. \(\displaystyle 1.6=\frac{y}{{1.5}}\) | 9. 7? тАУ 9 = 16 |

10. 14? тАУ 8 = 13 | 11. 17 + 6? = 9 | 12. \(\displaystyle \frac{x}{3}+1=\frac{7}{{15}}\) |

Solutions:

**1. ? тАУ 2 = 7**

Sol.: ? – 2 = 7

? = 7 + 2 (by transposing)

? = 9 answer

**2. ? + 3 = 10**

Sol.:┬а ? + 3 = 10

? = 10 – 3 (by transposing)

? = 7 answer

**3. 6 = ? + 2**

Sol.: 6 = ? + 2

or ? + 2 = 6

? = 6 – 2 *(by transposing)*

? = 4 answer

**4. \(\displaystyle \frac{3}{7}+x=\frac{{17}}{7}\)**

Sol.: \(\displaystyle \frac{3}{7}+x=\frac{{17}}{7}\)

\(\displaystyle x=\frac{{17}}{7}-\frac{3}{7}\) *(by transposing)*\(\displaystyle x=\frac{{17-3}}{7}\)

*(by L.C.M)*

\(\displaystyle x=\frac{{14}}{7}\)

? = 2 answer

**6. 6? = 12**

Sol.: 6? = 12

\(\displaystyle x=\frac{{12}}{6}\)

? = 2 answer

**7. \(\displaystyle \frac{{2x}}{3}=18\)**

Sol.: \(\displaystyle \frac{{2x}}{3}=18\)

\(\displaystyle \frac{{2x}}{3}=\frac{{18}}{1}\)

2? ├Ч 1 = 18 ├Ч 3 *(by cross multiplication)*2? = 54

\(\displaystyle \frac{{54}}{2}\)? = 27

*answer*

**8. \(\displaystyle 1.6=\frac{y}{{1.5}}\)**

Sol.: \(\displaystyle 1.6=\frac{y}{{1.5}}\)\(\displaystyle \frac{{1.6}}{1}=\frac{y}{{1.5}}\)? = 1.5 ├Ч 1.6

? = 2.4 answer

**9. 7? тАУ 9 = 16**

Sol.: 7? – 9 = 16

7? = 16 + 9

7? = 25

\(\displaystyle \frac{{25}}{7}\) answer

**10. 14? тАУ 8 = 13**

Sol.: 14? – 8 = 13

14? = 13 + 8

14? = 21

? = \(\displaystyle \frac{{21}}{{14}}\)

? = \(\displaystyle \frac{3}{2}\) answer

**11. 17 + 6? = 9**

Sol.: 17 + 6? = 9

6? = 9 – 17 (by transposing)

6? = – 8

\(\displaystyle \frac{{-8}}{6}\)

\(\displaystyle \frac{{-4}}{3}\) answer

**12. \(\displaystyle \frac{x}{3}+1=\frac{7}{{15}}\)**

Sol.: \(\displaystyle \frac{x}{3}+1=\frac{7}{{15}}\)

\(\displaystyle \frac{x}{3}=\frac{7}{{15}}-\frac{1}{1}\) (by transposing)

\(\displaystyle \frac{x}{3}=\frac{{7-1}}{{15}}\) (by L.C.M.)

\(\displaystyle \frac{x}{3}=\frac{6}{{15}}\)

15 ├Ч ? = 6 ├Ч 3 (by cross multiplication)

15? = 18

\(\displaystyle x=\frac{{18}}{{15}}\)

\(\displaystyle x=\frac{{6}}{{5}}\) answer

## Ncert Solutions for class 8 maths chapter 2 EXERCISE 2.2

class 8th maths

**1. If you subtract \(\displaystyle \frac{1}{2}\) from a number and multiply the result by \(\displaystyle \frac{1}{2}\), you get \(\displaystyle \frac{1}{8}\). What is the number?**

Sol.: Let’s the number is = ?

subtract \(\displaystyle \frac{1}{2}\) from x = \(\displaystyle x-\frac{1}{2}\)

Result is \(\displaystyle \frac{{2x-1}}{2}\)

Result multiply by \(\displaystyle \frac{1}{2}\)

According to the question –

\(\displaystyle \begin{array}{l}\left( {\frac{{2x-1}}{2}} \right)\times \frac{1}{2}=\frac{1}{8}\\\frac{{2x-1}}{4}=\frac{1}{8}\end{array}\)

\(\displaystyle 8\times (2x-1)=4\times 1\) (by cross multifaction)

16? – 8 = 4┬а

16? = 4 + 8 (by transposing)

16? = 12

\(\displaystyle \begin{array}{l}x=\frac{{12}}{{16}}\\x=\frac{3}{4}\end{array}\)

so that number is \(\displaystyle \frac{3}{4}\) answer.

**2. The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool?**

Sol.: Let’s the breadth of the swimming pool is = ?┬а

so Its length is = 2? + 2

Perimeter of a rectangle is = 154 m

2 (length + breadth) = 154

put the value –

2 (2? + 2 + ?) = 154

2 (3? +2) = 154

6? + 4 = 154

6? = 154 – 4

6? = 150

\(\displaystyle x=\frac{{150}}{6}\)

? = 25

so the breadth of the pool is ? = 25 m

and the length of the pool is 2? + 2┬а

2 ├Ч 25 + 2 = 52 m

**3. The base of an isosceles triangle is \(\displaystyle \frac{4}{3}\) cm. The perimeter of the triangle is \(\displaystyle 4\frac{2}{{15}}\) cm. What is the length of either of the remaining equal sides?**

Sol.: Let’s the equal side is = ?

Base of isosceles triangle is = \(\displaystyle \frac{4}{3}\) m

Perimeter of the triangle = \(\displaystyle 4\frac{2}{{15}}\)

Sum of all three sides = \(\displaystyle 4\frac{2}{{15}}\)

\(\displaystyle \begin{array}{l}x+x+\frac{4}{3}=4\frac{2}{{15}}\\2x+\frac{4}{3}=\frac{{62}}{{15}}\end{array}\)

\(\displaystyle 2x=\frac{{62}}{{15}}-\frac{4}{3}\) (by transposing)

\(\displaystyle \begin{array}{l}2x=\frac{{62-20}}{{15}}\\2x=\frac{{42}}{{15}}\\2x=\frac{{14}}{5}\\x=\frac{{14}}{5}\times \frac{1}{2}\\x=\frac{7}{5}\end{array}\)

The length of each of the remaining equal sides x = \(\displaystyle \frac{7}{5}\) cm

or \(\displaystyle 1\frac{2}{5}\) cm

**4. Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.**

Sol.: Let’s the first number is = ?

and the second number is = ? + 15

according to the question –

? + ? + 15 = 95

2? + 15 = 95

2? = 95 – 15 (by transposing)

2? = 80

? = \(\displaystyle \frac{{80}}{2}\)

? = 40

so the first number ? = 40 answer

and the second number is = ? + 15

= 40 + 15

= 55 answer

**5. Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?**

Sol.: Let’s first number is = 5?

and the second number is = 3?

according to the question –

5?- 3? = 18

2? = 18

? = \(\displaystyle \frac{{18}}{2}\)

? = 9

so the first number is 5? = 5 ├Ч 9

= 45 answer

and second number is 3? = 3 ├Ч 9

= 27 answer

**6. Three consecutive integers add up to 51. What are these integers?**

Sol.: Let’s first integers is = ?

second integers is = ? + 1 (difference of two consecutive integers)

so third integers is = ? + 2

according to the question –

? + ? + 1 + ? + 2 = 51

3? + 3 = 51

3? = 51 – 3 (by transposing)

3? = 48

? = \(\displaystyle \frac{{48}}{3}\)

? = 16

so the first integers ? = 16 answer

second integers is ? + 1 = 16 + 1 = 17 answer

third integers is ? + 2 = 16 + 2 = 18 answer

These three consecutive integers are 16, 17 and 18.

**7. The sum of three consecutive multiples of 8 is 888. Find the multiples.**

Sol.: Let’s first multiples of 8 is = ?

second multiple of 8 is = ? + 8 (difference of two consecutive multiples of 8)

and third multiple of 8 is = ? + 16

according to the question –

? + ? + 8 + ? + 16 = 888

3? + 24 = 888

3? = 888 – 24

3? = 864

? = \(\displaystyle \frac{{888}}{3}\)

? = 288

So the first multiple of 8 is ? = 288

second multiple of 8 is ? + 8 = 288 + 8 = 296

third multiple of 8 is ? + 16 = 288 + 16 = 304

The three consecutive multiples of 8 is 288, 296 and 304.

**8. Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.**

Sol.: Let’s first integers is = ?

Second integers is = ? + 1

and third integers is = ? + 2

According to the question –

Respectively multiplied by 2, 3 and 4.

2 ├Ч ? + 3 ├Ч (? + 1) + 4 ├Ч (? + 2) = 74

2? + 3? + 3 + 4? + 8 = 74

9? + 11 = 74

9? = 74 – 11 (by transposing)

9? = 63

? = \(\displaystyle \frac{{63}}{9}\)

? = 7

So first integers is ? = 7

Second integers is ? + 1 = 7 + 1 = 8

and third integers is ? + 2 =┬а 7 + 2 = 9

These three consecutive integers are 7. 8 and 9.

**9. The ages of Rahul and Haroon are in the ratio 5 : 7. Four years later the sum of their ages will be 56 years. What are their present ages?**

Sol.: Let’s Rahul age is = 5?

and Haroon age is = 7?

After 4 years Rahul age = 5? + 4

and Haroon age = 7? + 4

according to the question –

5? + 4 + 7? + 4 = 56

12? + 8 = 56

12? = 56 – 8 (by transposing)

12? = 48

? = \(\displaystyle \frac{{48}}{12}\)

? = 4

So, Rahul age is 5? = 5 ├Ч 4 = 20 year

and Haroon age is 7? = 7 ├Ч 4 = 28 year┬а

**10. The number of boys and girls in a class are in the ratio 7 : 5. The number of boys is 8 more than the number of girls. What is the total class strength?**

Sol.: Let’s the number of girls are = ?

and the number of boys are = ? + 8

according to the question –

\(\displaystyle \frac{{x+8}}{x}=\frac{7}{5}\)

\(\displaystyle 7\times x=5\times (x+8)\) (by cross multiplication)

\(\displaystyle \begin{array}{l}7x=5x+40\\7x-5x=40\\2x=40\end{array}\)

? = \(\displaystyle \frac{{40}}{2}\)

? = 20

So the number of girls are ? = 20

and the number of boys are ? + 8 = 20 + 8 = 28

the total class strength is 20 + 28 = 48 answer

**11. BaichungтАЩs father is 26 years younger than BaichungтАЩs grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?**

Sol.: Let’s Baichung father’s age is = ?

Baichung grandfather’s age = ? + 26

and Baichung’s age = ? – 29

according to the question –

? + ? + 26 + ? – 29 = 135

3? – 3 = 135

3? = 135 + 3 (by transposing)

3? = 138

? = \(\displaystyle \frac{{138}}{3}\)

? = 46

So, BaichungтАЩs father age ? = 46 years

Baichung’s grandfather age ? + 26

= 46 + 26 = 72 years

Baichung’s age ? – 29

= 46 – 29 = 17 years

**12. Fifteen years from now RaviтАЩs age will be four times his present age. What is RaviтАЩs present age?**

Sol.: Let’s Ravi’s present age is = ?

after 15 years Ravi’s age = ? + 15

according to the question –

? + 15 = 4?

? – 4x = – 15 (by transposing)

– 3? = – 15

? = \(\displaystyle \frac{{-15}}{-3}\)

? = 5

So Ravi’s present age is = 5 years

**13. A rational number is such that when you multiply it by \(\displaystyle \frac{5}{2}\) and add \(\displaystyle \frac{2}{3}\) to the product, you get \(\displaystyle -\frac{7}{{12}}\). What is the number?**

Sol.: Let’s the rational number is = ?

x is multiply by \(\displaystyle \frac{5}{2}\) then product is

\(\displaystyle x\times \frac{5}{2}\)

\(\displaystyle \frac{5x}{2}\)

adding \(\displaystyle \frac{2}{3}\) then

according to the question –

\(\displaystyle \frac{{5x}}{2}+\frac{2}{3}=-\frac{7}{{12}}\)

\(\displaystyle \frac{{15x+4}}{6}=-\frac{7}{{12}}\) (by L.C.M)

\(\displaystyle 12\times (15x+4)=-7\times 6\) (by cross multiplication)

\(\displaystyle \begin{array}{l}180x+48=-42\\180x=-42-48\\180x=-90\\x=\frac{{-90}}{{180}}\end{array}\)

\(\displaystyle x=\frac{{-1}}{2}\)

So rational number is \(\displaystyle \frac{1}{2}\) answer

**14. Lakshmi is a cashier in a bank. She has currency notes of denominations тВ╣ 100, тВ╣ 50 and тВ╣ 10, respectively. The ratio of the number of these notes is 2 : 3 : 5. The total cash with Lakshmi is 4,00,000. How many notes of each denomination does she have?**

Sol.: тИ╡ Ratio of тВ╣ 100 notes is = 2

Let’s number of тВ╣ 100 notes = 2?

So, Value of тВ╣ 100 notes = 2? ├Ч 100 = 200?┬а

тИ╡ Ratio of тВ╣ 50 notes is = 3

Let’s number of тВ╣ 50 notes = 3?

So, Value of тВ╣ 50 notes = 3? ├Ч 50 = 150?

тИ╡ Ratio of тВ╣ 10 notes is = 5

Let’s number of тВ╣ 10 notes = 5?

So, Value of тВ╣ 10 notes = 5? ├Ч 10 = 50?┬а

according to the question –┬а

200? + 150? + 50? = 400000

400? = 400000

\(\displaystyle x=\frac{{400000}}{{400}}\)

? = 1000

So the number of тВ╣100 notes = 2? = 2 ├Ч 1000 = 2000 notes

number of тВ╣ 50 notes = 3? = 3 ├Ч 1000 = 3000 notes

number of тВ╣ 10 notes = 5? = 5 ├Ч 1000 = 5000 notes

**15. I have a total of тВ╣ 300 in coins of denomination тВ╣ 1, тВ╣ 2 and тВ╣ 5. The number of тВ╣ 2 coins is 3 times the number of тВ╣ 5 coins. The total number of coins is 160. How many coins of each denomination are with me?**

Sol.: Let’s number of coins тВ╣ 5 is = *x*

and number of coins тВ╣ 2 is = 3x

Total number of coins is = 160

Then number of coins тВ╣ 1 is = (160 – 3x – x) = (160-4x)

Value of тВ╣ 1 coins = (160 – 4x) ├Ч 1 = 160 – 4x

Value of тВ╣ 2 coins = 3x ├Ч 2 = 6x

Value of тВ╣ 5 coins = x ├Ч 5 = 5x

According to the question –

160 – 4x + 6x + 5x = 300

160 – 4x + 11x = 300

7x = 300 – 160 (by transposing)

7x = 140

\(\displaystyle x=\frac{{140}}{{7}}\)

x = 20

So, Number of тВ╣ 5 coins is = x

= 20 coins

Number of тВ╣ 2 coins is = 3x

= 3 ├Ч 20

= 60 coins

Number of тВ╣ 1 coins is = 160 – 4x

= 160 – 4 ├Ч 20

= 160 – 80┬а

= 80 coins

**16. The organisers of an essay competition decide that a winner in the competition gets a prize of тВ╣ 100 and a participant who does not win gets a prize of тВ╣ 25. The total prize money distributed is тВ╣ 3,000. Find the number of winners, if the total number of participants is 63.**

Sol.: Total number of participants is = 63

Let’s number of winner is = x

and number of participant who does not win = 63 – x

Each winner gets a prize of = тВ╣ 100

So total winner prize money = x ├Ч 100

= 100x

Same, each participant who does not win gets a prize of = тВ╣ 25

So total each participant who does not win gets a prize money = (63 – x) ├Ч 25

= 1575 – 25x

According to the question –

100x + 1575 – 25x = 3000

75x = 3000 – 1575 (by transposing)

75x = 1425

\(\displaystyle x=\frac{{1575}}{{75}}\)

x = 19

So the number of winners is 19 answer.

## Ncert Solutions for Class 8 Maths Chapter 2 EXERCISE 2.3

class 8th maths

**Solve the following equations and check your results.**

1. 3x = 2x + 18 | 2. 5t тАУ 3 = 3t тАУ 5 | 3. 5x + 9 = 5 + 3x |

4. 4z + 3 = 6 + 2z | 5. 2x тАУ 1 = 14 тАУ x | 6. 8x + 4 = 3 (x тАУ 1) + 7 |

7. \(\displaystyle x=\frac{4}{5}(x+10)\) | 8. \(\displaystyle \frac{{2x}}{3}+1=\frac{{7x}}{{15}}+3\) | 9. \(\displaystyle 2y+\frac{5}{3}=\frac{{26}}{3}-y\) |

10. \(\displaystyle 3m=5m-\frac{8}{5}\) | ┬а | ┬а |

**Solutions :**

**1. 3x = 2x + 18**

Sol.: 3x = 2x + 18

3x – 2x = 18 (by transposing)

x = 18 answer┬а

**2. 5t тАУ 3 = 3t тАУ 5**

Sol.: 5t – 3 = 3t – 5

5t – 3t = – 5 + 3

2t = – 2

\(\displaystyle t=-\frac{2}{2}\)

t = – 1 answer┬а

**3. 5x + 9 = 5 + 3x**

Sol.: 5x + 9 = 5 + 3x

5x – 3x = 5 – 9 (by transposing)

2x = – 4

\(\displaystyle x=-\frac{4}{2}\)

x = – 2 answer┬а

**4. 4z + 3 = 6 + 2z**

Sol.: 4z + 3 = 6 + 2z

4z – 2z = 6 – 3 (by transposing)

2z = 3

\(\displaystyle z=\frac{{3}}{{2}}\) answer

**5. 2x тАУ 1 = 14 тАУ x**

Sol.: 2x – 1 = 14 – x

2x + x = 14 + 1 (by transposing)

3x = 15

\(\displaystyle x=\frac{{15}}{{3}}\)

x = 5

**6. 8x + 4 = 3 (x тАУ 1) + 7**

Sol.: 8x + 4 = 3 (x – 1) + 7

8x + 4 = 3x – 3 + 7

8x – 3x = – 3 + 7 –┬а 4 (by transposing)

5x = 7 – 7

5x = 0

\(\displaystyle x=\frac{{0}}{{5}}\)

x = 0 answer

**7. \(\displaystyle x=\frac{4}{5}(x+10)\)**

Sol.: \(\displaystyle x=\frac{4}{5}(x+10)\)

\(\displaystyle \begin{array}{l}x=\frac{{4x}}{5}+\frac{4}{5}\times 10\\x=\frac{{4x}}{5}+\frac{8}{1}\end{array}\)

\(\displaystyle x=\frac{{4x+40}}{5}\) (by L.C.M.)

\(\displaystyle 5x=4x+40\)

\(\displaystyle 5x-4x=40\) (by transposing)

x = 40 answer

**8. \(\displaystyle \frac{{2x}}{3}+1=\frac{{7x}}{{15}}+3\)**

Sol.: \(\displaystyle \frac{{2x}}{3}+1=\frac{{7x}}{{15}}+3\)

\(\displaystyle \frac{{2x}}{3}-\frac{{7x}}{{15}}=3-1\) (by transposing)

\(\displaystyle \frac{{10x-7x}}{{15}}=\frac{2}{1}\) (by cross multiplication)

10x – 7x = 2 ├Ч 15

3x = 30

\(\displaystyle x=\frac{{30}}{3}\)

x = 10 answer

**9. \(\displaystyle 2y+\frac{5}{3}=\frac{{26}}{3}-y\)**

Sol.: \(\displaystyle 2y+\frac{5}{3}=\frac{{26}}{3}-y\)

\(\displaystyle 2y+y=\frac{{26}}{3}-\frac{5}{3}\)┬а

\(\displaystyle \begin{array}{l}3y=\frac{{26-5}}{3}\\3y=\frac{{21}}{3}\end{array}\) (by transposing)

\(\displaystyle y=\frac{7}{3}\) answer

**10. \(\displaystyle 3m=5m-\frac{8}{5}\)**

Sol.: \(\displaystyle 3m=5m-\frac{8}{5}\)

\(\displaystyle 3m-5m=-\frac{8}{5}\) (by transposing)

\(\displaystyle -2m=-\frac{8}{5}\) (both side have – sign so – to – cancel)

\(\displaystyle m=\frac{8}{{2\times 5}}\)

\(\displaystyle m=\frac{4}{5}\) answer